From the book "Calculus" by James Stewart,

The Closed Interval Method is used to find the absolute(global) maximum and minimum values of a continuous function on a close interval $[a,b]$.

1) Find the values of $f$ at the critical numbers of $f$ in $(a,b)$.

2) Find the values of $f$ at the endpoints of the interval.

3) The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.

I have problem understanding step 1.

Fermat's Theorem states that if function $f$ has local maximum/minimum at point $x=a$, then $a$ is a critical number.

Also the converse is not true as the function $y=x^3$ has a critical point at $x=0$ but $x=0$ is not a local minimum/maximum.

Also, from my understanding, absolute minimum/maximum on an closed interval can either be a local min/max on the closed interval or at the end points.

Hence, we should find the local minimum/maximum and compare the values with the value of function at the endpoints.

But from the example of $y=x^3$, doing step 1) as suggested in the book can derive point that is a critical number but not a local minimum/maximum.

Perhaps, it is proven such points can never be the absolute maximum/minimum?

Edit: knowing such point $a$ is not a local minimum/maximum tells us that there exist $x$ such that $f(x) > a$ and $f(x) < a$ and the absolute maximum/minimum would either be another critical number or at the end points.


2 Answers 2


Step 1 is just part of the whole story. And you are right that Step 1 would give points that are not absolute min/max. That's exactly why you need Step 2 and Step 3.

In the case of $f(x)=x^3$ (say in $[-1,1]$), it is true that you will get $x=0$ as a critical number. However, Step 2 and Step 3 will rule it out.

[Added to answer the question in the comment:] Because global min/max must also be local min/max. If a critical point is not a local min/max, then it cannot be global min/max.

  • $\begingroup$ why is it that Step 2 and Step 3 will definitely rule out such points? Can you explain in more details? $\endgroup$ Commented Jan 4, 2017 at 15:21
  • $\begingroup$ In step 1, you get $f(0)=0$. But in Step 2, you get $f(-1)=-1$ and $f(1)=1$. Now by step 3, what you get? $\endgroup$
    – user9464
    Commented Jan 4, 2017 at 15:24
  • $\begingroup$ How is it that step 2 and step 3 will rule out such points in general? Not only in the case of $y=x^3$ $\endgroup$ Commented Jan 4, 2017 at 15:28
  • $\begingroup$ It is because by definition so. In Step 3, one compares all the points one gets in Step 1 and Step 2 so that one can identify the absolute max and absolute min. The bad points you described would not "win" in the comparison because they cannot be the absolute max/min. $\endgroup$
    – user9464
    Commented Jan 4, 2017 at 15:31
  • $\begingroup$ Intuitively, they cannot be the absolute maximum/minimum. But is there a formal proof that i can refer to? $\endgroup$ Commented Jan 4, 2017 at 15:53

Just a thought, since the end points of a closed interval function say [a,b] can be a local extremum and at the same time, Fermat's theorem says that all local extremums occur at critical points. Does it mean to say that x=a and x=b is a critical point? For Eg, f(x)=x^3 where x is defined in the interval [-3,4]. x=-3 and x=4 are both local extremums but why are they considered critical points? f'(a) and f'(b) is not zero and is defined. The definition of a critical point is such that if x=c is a critical point, f'(c) is either 0 or undefined.


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